Derivative of Square Root of x

Leave a Comment / Mathematics / By Azhar

$\frac12x^{-\frac12}$ is the derivative of square root of $x$ . Different methods of differentiation can be used to calculate this derivative, such as the chain rule, the power rule, and the first principle of derivatives. A derivative of square root of x can be written mathematically as $\frac d{dx}\sqrt x=\frac12x^{-\frac12}$ or $=\frac1{2\sqrt x}$ , using the power rule $\frac d{dx}x^n=nx^{n-1}$ . The derivative of the square root of x can be obtained by using this formula and substituting $n = {\frac12}$ .

What is the Derivative of the Square Root of x?

The derivative of square root of x is, $\frac12x^{-\frac12}$ . We know, that in mathematics the derivative of a function is the process to find out the rate of change of a function with respect to a variable. We can find the derivative of root x by two methods

Power rule of differentiation
By the first Principle of differentiation

Derivative of Square Root of x Using Power Rule:

Power Rule Formula for derivative is:

$\frac d{d_x}x^n=nx^{n-1},$ .

Square Root of x is an exponential function with x as the base and $n = {\frac12}$ as the power.

Now,

If substitute $n = {\frac12}$ in the formula

$\frac d{dx}x^n=nx^{n-1}$ ,

then

$\frac d{dx}x^\frac12=\frac12x^{\frac12-1}$ .

$\frac d{dx}x^\frac12=\frac12x^{-\frac12}$ .

$\frac d{dx}x^\frac12 =\frac1{2\sqrt x}$ .

Hence, Proved the derivative of square root of x is equal to $=\frac1{2\sqrt x}$ .

Derivative of square Root of x Using First Principle:

$f(x)\; = \;\sqrt x$

$f'(x)\;=\;\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}h$ .

$f'(x)\;=\;\lim_{h\rightarrow0}\frac{\sqrt{x+h}-\sqrt x}h$ .

$f'(x)\;=\;\lim_{h\rightarrow0}\frac{\sqrt{x+h}-\sqrt x}h\times\frac{\sqrt{x+h}\;+\sqrt x}{\sqrt{x+h}\;+\sqrt x}$ .

$f'(x)\;=\;\lim_{h\rightarrow0}\frac{(x+h)-x}{\sqrt{x+h}+\sqrt x}$ .

$f'(x)\;=\;\lim_{h\rightarrow0}\frac h{h(\sqrt{x+h}+\sqrt x}$ .

$f'(x)\;=\;\lim_{h\rightarrow0}\frac1{(\sqrt{x+h}+\sqrt x}$ .

$f'(x)\;=\;\frac1{(\sqrt{x+0}+\sqrt x}$ .

$f'(x)\;=\;\frac1{2\sqrt x}$ .

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